Since they're both multiplied by $(n+1)^2$, it's easy to see that if $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$, then $(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$. But if the $\ge$ were replaced with a $\le$, the proof would still be valid, even though it's demonstrably not true.

I can see that it would take strong induction would fix this problem, but if I didn't know by observation that $\left(\sum_{k=1}^nk\right)^2\le\sum_{k=1}^nk^2$ isn't true by observation, how would I know to use strong induction?

3 Answers
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You're not quite on the right track. Your base case is just fine. For your induction step, suppose that $$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2.$$ Note then that $$\left(\sum_{k=1}^{n+1}k\right)^2=\left((n+1)+\sum_{k=0}^nk\right)^2=(n+1)^2+2(n+1)\sum_{k=1}^nk+\left(\sum_{k=1}^nk\right)^2$$ and that $$\sum_{k=1}^{n+1}k^2=(n+1)^2+\sum_{k=1}^nk^2.$$ Can you get the rest of the way from there?